The home page presentation provides insight into the operation of two ideal LC circuits: A seriesresonant circuit (without a series resistor), and a parallelresonant circuit (without a parallel resistor). The author asserts that determining solution sets for ideal circuit configurations should precede attempts to analyze their parasiticladen counterparts. Gaining an intuitive understanding of ideal circuit operation ought to be a high priority. Then, when parasitics are introduced, the practitioner will be better prepared to understand what is observed and, if necessary, synthesize what is needed for some particular application.
Circuit 1 (a seriesresonant configuration)
Initial Conditions
1. The charge on the capacitor is 0. Therefore its voltage, V_{C}, is 0V 2. No current is flowing in the loop, nor is the current changing.
The Circuit Elements
• The battery is ideal (no resistance, no inductance, and the voltage is invariant with load) • The switch is ideal (∞ Ohms when open, 0 Ohms when closed, no inductance, no bounce) • The inductor is ideal (no resistance, no capacitance, and L is a fixed value) • The capacitor is ideal (no resistance, no inductance, and C is a fixed value) • The diode is ideal (no forward voltage drop, infinite reverse breakdown) • The wire is ideal (no resistance, no inductance)
The Event
At t=0 the switch closes, and remains closed
The Tasks
1. Determine an expression for current in the loop. 2. Determine an expression for voltage on the inductor. 3. Determine an expression for voltage on the capacitor.
Observations
• ..No current will be flowing just after the switch is thrown because current cannot change instantaneously in an inductor. • The voltage will be 0V on the capacitor just after the switch is thrown because current has not begun flowing. • Therefore all the battery voltage will appear across L just after the switch is thrown. • The slope of V_{C} just after the switch is thrown must be zero because i=C(dv/dt), and i=0.
Note
Readers familiar with switching regulators should recognize Figure 1 as an idealized building block. The diode is placed to manage the current flowing in the inductor when the switch opens (because current cannot change instantaneously in an inductor). In addition, there would exist some feedback mechanism to control the switch in response to demands placed on the load. Highly efficient regulators would specify particular diode types, or they may even replace the diode with some synchronized switch circuitry. The form this synchronizing circuitry might take is not a subject of this work, nor is the type of diode, nor the switch structure.
The Analytic Solution
If we write an equation for the voltage around the loop after the switch has closed, we get
.......V_{Batt}  L(di/dt)  (1/C)∫idt = 0
Or, using an abbreviated notation, we have
............V_{Batt}  Li'  (1/C)∫idt = 0
If we differentiate with respect to time, we see the following
....................0  Li''  i/C = 0
Multiplying by 1, and dividing by L yields a classic 2ndorder differential equation
......................i'' + i/(LC) = 0
The general solution is given by
.................................i = Asinωt + Bcosωt
Where
.................................ω = 1/SQRT(LC)
and where we note that
.................................ω = 2πf
At time t=0 i is 0. Therefore, since sin0=0 and cos0=1, B must be 0, and we can write
.................................i = Asinωt
We note that
................................v_{L} = Ldi/dt
Then
................................v_{L} = LAωcosωt
At t=0 V_{L} equals V_{Batt}. The cosine of 0 is 1. So, we have
..............................V_{Batt} = LAω
Solving for A, we get
.................................A = V_{Batt}/(Lω)
Substituting 1/SQRT(LC) for ω permits A to be written even more precisely as
.................................A = V_{Batt}(SQRT(C/L))
which is the peak current. The complete expression for i is then given by
.................................i = V_{Batt}(SQRT(C/L))sinωt
The peak current occurs when sinωt is 1. This occurs many times (because the circuit is oscillating). The first time it occurs is when ωt equals π/2. Solving for t, we have the following expression for the time of the first peak current
.................................t = π/2ω
Recalling that ω = 1/SQRT(LC), we can rewrite the above equation, to give
.................................t = π(SQRT(LC))/2
which is the first occurrence of the peak current.
The voltage on the inductor is given by Ldi/dt. Thus (with a modicum of algebraic manipulation), we arrive at
................................v_{L} = LV_{Batt}(SQRT(C/L))ωcosωt
Then, substituting 1/SQRT(LC) for ω results in the following compact expression
................................v_{L} = V_{Batt}cosωt
Now, what about the peak voltage on the capacitor? Well,
................................v_{C} = (1/C)∫idt
Doing the integration produces the following result:
................................v_{C} = V_{Batt}(1/C)(SQRT(C/L))((1/ω)cosωt + k)
Where k is a constant of integration. We have established that v_{C}=0 at t=0. Observing that cos0=1 leads us to the conclusion that k must be given by
.................................k = 1/ω
Gathering and rearranging terms allows us to write
................................v_{C} = V_{Batt}(1/C)(SQRT(C/L))(1/ω)(1cosωt)
But ω=1/SQRT(LC). Substituting this in, and then grinding through the math, gives the following succinct relationship
................................v_{C} = V_{Batt}(1cosωt)
An easier way to get this solution would be to apply Kirkhoff's voltage law on the loop, as follows:
................................v_{C} = V_{Batt}  v_{L} Knowing that v_{L}=V_{Batt}cosωt leads to an identical expression for v_{C} (as it should).
cosωt ranges from +1 to 1. Therefore v_{C} ranges from 0 to 2V_{Batt} (indicating that this circuit is a voltage doubler). And when does it peak at 2V_{Batt}? That occurs whenever cosωt=1, which occurs many times (because, again, this circuit is oscillating). But the first time it occurs is when the following relationship is satisfied
................................ωt = π
Solving for t, we get .................................t = π/ω
which is twice the time it takes for the current to peak. Thus
.................................t = π(SQRT(LC))
which is the first occurrence of peak voltage on the capacitor.
Solution Summary for the Case of Zerovolt and Zerocurrent Initial Conditions
..............................i_{Loop} = V_{Batt}(SQRT(C/L))sinωt
................................v_{L} = V_{Batt}cosωt
................................v_{C} = V_{Batt}(1cosωt)
where .................................ω = 1/SQRT(LC)
Implications for Stepdown Switching Regulators
Ordinarily the switch is opening & closing (regulating the output voltage to some level well below V_{Batt}). But suppose (because of some circuit failure) it initially closes, and then does not reopen. The output will rise to twice the battery voltage. This could be catastrophic for the load, and must therefore be mitigated.
A Note About Series Resistance
With the possible exception of circuit operation near Absolute Zero (0°K), there are no known zeroresistance circuits. The switch (usually a MOSFET) and the inductor would be likely candidates for introducing enough series resistance to merit some investigation. It should be apparent that adding a series resistance will introduce an additional term in the loop equation. Interested readers are referred to Series RLC Circuitry for an analytical treatment of this condition.
The More General Case
Although closing the switch when v_{C}=0, and when i_{Loop}=0 is an important case (i.e., startup for a switching regulator), two (likely simultaneous) initial conditions should also be considered. Those initial conditions are:
 A preexisting loop current, I_{0} (flowing in L, C, and through the diode)
 A preexisting voltage on the capacitor, V_{C0}
Recall that a regulated system has the switch opening & closing. We now define t=0 to be the moment just after the n^{th} closure of the switch. We assume again that some anamolous condition removes the switch from its usual controlled state of operation. That is, once closed it cannot reopen. Then we look again at Figure 1. We notice first that the diode is removed from consideration just as soon as it is forced to be in parallel with the battery. Current in the inductor cannot change instantaneously; therefore I_{0} continues to flow in L & C, but is diverted entirely to the battery. If we follow the original analysis it becomes apparent that we should now step in where, as before,
.................................i = Asinωt + Bcosωt
If we set t equal to 0 it becomes apparent that B=I_{0} (because the sine of 0 is 0, and the cosine of 0 is 1)
We note (again) that ................................v_{L} = Ldi/dt Then ................................v_{L} = LAωcosωt  Bωsinωt
Noting that v_{C}(t=0) equals V_{C0 }(the defined initial condition), we can write
...........................v_{L}(t=0) = V_{Batt}V_{C0}
And, because the cosine of 0 is 1, and the sine of 0 is 0, we can write
...........................V_{Batt}V_{C0} = LAω
A is now perfectly defined. .................................A = (V_{Batt}V_{C0})/Lω
Substituting 1/SQRT(LC) for ω permits A to be written even more precisely as
.................................A = (V_{Batt}V_{C0})(SQRT(C/L))
So, we now have a complete solution for i using only the applied voltage, circuit values, and the given initial conditions.
.................................i = (V_{Batt}V_{C0})(SQRT(C/L))sinωt + I_{0}cosωt
As before, the voltage on the inductor is given by Ldi/dt. So, we can write
................................v_{L} = L((V_{Batt}V_{C0})(SQRT(C/L))ωcosωt  I_{0}ωsinωt)
Expanding ω to 1/SQRT(LC), we can simplify v_{L} to the following form
................................v_{L} = (V_{Batt}V_{C0})cosωt  I_{0}SQRT(L/C)sinωt
At this point we express v_{C} as the battery voltage less the inductor voltage, as follows
................................v_{C} = V_{Batt}  (V_{Batt}V_{C0})cosωt + I_{0}SQRT(L/C)sinωt
Rearranging gives ................................v_{C} = V_{Batt}(1cosωt) + V_{C0}cosωt + I_{0}SQRT(L/C)sinωt
If (as a sanity check) we set V_{C0} to 0 and I_{0} to 0 we get precisely the same expression for v_{C} that derives from the startup case (as we should). So, we now have a general solution summary for all cases.
Solution Summary for the General Case of Nonzero Initial Conditions
..............................i_{Loop} = (V_{Batt}V_{C0})(SQRT(C/L))sinωt + I_{0}cosωt
................................v_{L} = (V_{Batt}V_{C0})cosωt  I_{0}SQRT(L/C)sinωt
................................v_{C} = V_{Batt}(1cosωt) + V_{C0}cosωt + I_{0}SQRT(L/C)sinωt
where .................................ω = 1/SQRT(LC)
Circuit 2 (a parallelresonant configuration)
Three ideal circuit elements in parallel:
Initial Conditions
The charge on the capacitor is 0. Therefore its voltage, V_{C}, is 0V No current is flowing anywhere in the network, nor is any current changing
The Circuit Elements
The current source is ideal (infinite impedance, no capacitance, and the current is invariant with voltage) The inductor is ideal (no resistance, no capacitance, and L is a fixed value) The capacitor is ideal (no resistance, no inductance, and C is a fixed value) The wire is ideal (no resistance)
The Event
At t=0 the current source switches from 0A to I.
The Tasks
Determine an expression for the voltage across the network. Determine an expression for current in the capacitor. Determine an expression for current in the inductor.
Observations
The voltage will be 0V just after the current steps to I because voltage cannot change instantaneously on a capacitor. i_{L} = 0 just after the current steps to I because current cannot change instantaneously in an inductor. i_{C} = I just after the current steps to I because no current is flowing in the inductor. Even though the magnitude of V_{C} at t=0 is 0V, the slope of V_{C }at t=0 must be >0 because i_{C}=C(dv/dt), and i is I.
The Analytic Solution
Kirkhoff's current law must be satisfied (the sum of the currents into any node must be zero). Thus,
.......................I  i_{L}  i_{C} = 0
Recognizing that i_{C}=Cdv/dt, and then rewriting, we get
...................I  i_{L}  Cdv/dt = 0
Differentiating with respect to time, we have
.................0  di_{L}/dt  Cv'' = 0
But Ldi_{L}/dt is the voltage on the inductor, which is the same as the voltage on the capacitor, which is v. So, we can write
....................0  v/L  Cv'' = 0
Multiplying through by 1, dividing by C, and then rearranging, gives
......................v'' + v/(LC) = 0
In form, this is identical to the switchedvoltage case. Therefore, the solutions for v, i_{C}, and i_{L} will be sinusoidal, and the analysis proceeds in a manner similar to the seriesresonant case. Without going through it stepbystep, the solution summary is given below.
Solution Summary for the Case of Zerovolt and Zerocurrent Initial Conditions
...............v_{CurrentSource} = v_{L} = v_{C} = I(SQRT(L/C))sinωt
................................i_{C} = Icosωt
................................i_{L} = I(1cosωt)
where
.................................ω = 1/SQRT(LC)
Implications
The current peak will be 2I in the inductor. This circuit may therefore be referred to as a current doubler. For that to predictably occur the current source must not saturate (if structured from a bipolar device) or go into the linear region of operation (if structured from a MOSFET). Therefore the peak voltage must be controlled by managing I, L, and C.
A Note About Parallel Resistance
With the possible exception of circuit operation near Absolute Zero (0°K), there are no known zeroresistance circuits. It should be apparent that adding a parallel resistance will introduce an additional term in the Kirkhoff equation. Interested readers are referred to Parallel RLC Circuitry for an analytical treatment of this condition.
What if the Input Source Does Not Differentiate to 0?
It may be that the input source is not a simple voltage or current step (each of which nicely differentiates to 0 for t>0). If that is the case, an appeal to one of the References will lead to the concept of the "particular" integral (solution). Or, applying an online search engine to one of the following strings may lead an investigator to a circuit with the input source of interest.
• LC Circuits • RLC Circuits
